2. A cylindrical tank 5 ft high has a constant diameter of 4 ft is full of water. The tank has a hole in its bottom that measures 0.1 ft². All losses are insignificant. How long will it take for the tank to empty? Note that you can NOT assume a static head and constant flow rate because the water head in the tank decreases as the water flows out of the tank.

Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter7: Analysis Of Stress And Strain
Section: Chapter Questions
Problem 7.7.10P: Solve the preceding problem for the following data: x=190106,y=230106,xy=160106,and=45.
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2. A cylindrical tank 5 ft high has a constant diameter of 4 ft is full of water. The tank has a hole
in its bottom that measures 0.1 ft². All losses are insignificant. How long will it take for the tank
to empty? Note that you can NOT assume a static head and constant flow rate because the
water head in the tank decreases as the water flows out of the tank.
(A) 35 sec
(B) 70 sec
(C) 105 sec
(D) 140 sec
Transcribed Image Text:2. A cylindrical tank 5 ft high has a constant diameter of 4 ft is full of water. The tank has a hole in its bottom that measures 0.1 ft². All losses are insignificant. How long will it take for the tank to empty? Note that you can NOT assume a static head and constant flow rate because the water head in the tank decreases as the water flows out of the tank. (A) 35 sec (B) 70 sec (C) 105 sec (D) 140 sec
From the conservation of mass.
min-mout=mcv
=mcv
0-m
mout
out
= -mcv
pxaxV=-px
Where, V = √ (2gH)
d
ax√(2gH):
dt 4
axV (2g) XH1/
ax√(2g) xdt=
After integrating,
ax√ (29) × /dt = - xD²×₁
11
x
==
d (7 ×D² xH)
==
-7
t = 70.0296 s
t≈70 s
t=70 s
π
XD² xH)
품
0.1 x√ (2x32.2) xt=
XD² x
XD² x.
(2g) xt=
XD²
Putting the known values
dH
H1/2
π
dH
dt
ex√ (29) Xt=-=xD³² x [01/2-1/2]
--
H1/2
1/2
میں
O
dH
H H1/2
51/2
·x4² x
x
1/2
Transcribed Image Text:From the conservation of mass. min-mout=mcv =mcv 0-m mout out = -mcv pxaxV=-px Where, V = √ (2gH) d ax√(2gH): dt 4 axV (2g) XH1/ ax√(2g) xdt= After integrating, ax√ (29) × /dt = - xD²×₁ 11 x == d (7 ×D² xH) == -7 t = 70.0296 s t≈70 s t=70 s π XD² xH) 품 0.1 x√ (2x32.2) xt= XD² x XD² x. (2g) xt= XD² Putting the known values dH H1/2 π dH dt ex√ (29) Xt=-=xD³² x [01/2-1/2] -- H1/2 1/2 میں O dH H H1/2 51/2 ·x4² x x 1/2
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